Area of a region R

Question

For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all the points $P$ lying in the first quadrant of the plane and satisfying $$2\le d_1(P) + d_2(P) \le 4$$ I am unable to understand how to proceed to solve the question. If the point was not situated on a plane would there be any change in the answer?

Answer 1

The first diagram shows the coordinate system with the graphs of $x+y=0$ and $x-y=0$ sketched as dashed lines. The polygon $ABCD$ is a square centered at the origin with sides parallel to the coordinate axes and the point $P$ is a point on the square lying a positive distance $d_1$ from the line $x+y=0$ and a positive distance $d_2$ from the line $x-y=0$.

Proposition: The area of square $ABCD$ is given by $\mathscr{A}=2(d_1+d_2)^2$

Proof:

1. $PB=\sqrt{2}d_1$

2. $AP=\sqrt{2}d_2$

3. $AB=AP+PB=\sqrt{2}(d_1+d_2)$

4. $\mathscr{A}=2(d_1+d_2)^2$

The second diagram shows two concentric squares centered at the origin with sides parallel to the axes.

The outer square has sides of length $4\sqrt{2}$ and the inner square has sides of length $2\sqrt{2}$.

From the proposition above it follows that for any point $P$ lying on the outer square, $d_1+d_2=4$ and that for any point $P$ lying on the inner square, $d_1+d_2=2$. For any point $P$ lying on or outside the inner square but on or inside the outer square, it will be the case that

$$ 2\le d_1+d_2\le 4 $$

Furthermore, the area of the region lying outside the smaller square but inside the larger square will be the difference of their areas:

\begin{eqnarray} \mathscr{A}_{outer}-\mathscr{A}_{inner}&=&2(4)^2-2(2)^2\\ &=&24 \end{eqnarray}

Since the problem asks only the area of the portion lying within the first quadrant, indicated by the polygon $R=LKDJIH$, the correct answer is that the area of $R$ is 6.